3x+x^2=160

Solution for 3x+x^2=160 equation:

3x+x^2=160

We move all terms to the left:

3x+x^2-(160)=0

a = 1; b = 3; c = -160;
Δ = b2-4ac
Δ = 32-4·1·(-160)
Δ = 649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{649}}{2*1}=\frac{-3-\sqrt{649}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{649}}{2*1}=\frac{-3+\sqrt{649}}{2}
$